aa r X i v : . [ m a t h . N T ] J u l On van Hamme’s (A.2) and (H.2) supercongruences
Ji-Cai Liu
Department of Mathematics, Wenzhou University, Wenzhou 325035, PR China [email protected]
Abstract.
In 1997, van Hamme conjectured 13 Ramanujan-type supercongruences la-beled (A.2)–(M.2). Using some combinatorial identities discovered by
Sigma , we extend(A.2) and (H.2) to supercongruences modulo p for primes p ≡ Keywords : Supercongruences; Hypergeometric series; p -Adic Gamma functions MR Subject Classifications : 05A19, 11A07, 33C20
In 1913, Ramanujan announced the following identity in his famous letter: ∞ X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k k ! ! = 2Γ (cid:0) (cid:1) , (1.1)where ( a ) = 1 and ( a ) n = a ( a + 1) · · · ( a + n −
1) for n ≥
1. The formula (1.1) was laterproved by Hardy (1924) and Watson (1931).Motivated by some of such formulas, van Hamme [13] conjectured 13 supercongruencesin 1997, which relate the partial sums of certain hypergeometric series to the values of the p -adic Gamma functions. These 13 conjectural supercongruences labeled (A.2)–(M.2) are p -adic analogues of their corresponding infinite series identities. For instance, van Hammeconjectured that the identity (1.1) has the following interesting p -adic analogue, whichwas first proved by McCarthy and Osburn [5] by use of Gaussian hypergeometric series. Theorem 1.1 (See [13, (A.2)].) For any odd prime p , we have p − X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k k ! ! ≡ − p Γ p (cid:0) (cid:1) (mod p ) if p ≡ , p ) if p ≡ ,where Γ p ( · ) denotes the p -adic Gamma function recalled in the next section. Later, Swisher [12] showed that the supercongruence (A.2) also holds modulo p forprimes p ≡ p ≡ p extension in terms of the p -adic Gamma functions. Interestingly, numericalcalculation suggests the following result which partially motivates this paper.1 onjecture 1.2 For primes p ≥ with p ≡ , we have p − X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k k ! ! ≡ − p
16 Γ p (cid:18) (cid:19) (mod p ) . (1.2)It is very difficult for us to prove (1.2) thoroughly, but we can show that (1.2) holdsmodulo p . Theorem 1.3
Let p ≥ be a prime. For p ≡ , we have p − X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k k ! ! ≡ − p
16 Γ p (cid:18) (cid:19) (mod p ) . (1.3)The other motivating example of this paper is the following supercongruence whichwas both conjectured and confirmed by van Hamme [13]. Theorem 1.4 (See [13, (H.2)].) For any odd prime p , we have p − X k =0 (cid:0) (cid:1) k k ! ! ≡ − Γ p (cid:0) (cid:1) (mod p ) if p ≡ , p ) if p ≡ . Recently, Long and Ramakrishna [4, Theorem 3] obtained a modulo p extension of(H.2) as follows: p − X k =0 (cid:0) (cid:1) k k ! ! ≡ − Γ p (cid:0) (cid:1) (mod p ) if p ≡ − p Γ p (cid:0) (cid:1) (mod p ) if p ≡ p extension of (1.4) for primes p ≡ Theorem 1.5
Let p ≥ be a prime. For p ≡ , we have p − X k =0 (cid:0) (cid:1) k k ! ! ≡ − p p (cid:18) (cid:19) J p − / (mod p ) , (1.5) where J n = P nk =0 (cid:0) kk (cid:1) / k . Remark.
We shall show that J ( p − / ≡ − (mod p ), which implies (1.5) is indeed amodulo p extension of the second case of (1.4). By the Chu-Vandermonde identity, wehave n X k =0 (cid:18) n + 1 k (cid:19) = 12 n +1 X k =0 (cid:18) n + 1 k (cid:19) = 12 (cid:18) n + 22 n + 1 (cid:19) . n = p − with p ≡ (cid:18) p − k (cid:19) ≡ (cid:0) kk (cid:1) ( − k (mod p ) and (cid:18) p − p − (cid:19) ≡ − p ) , we immediately obtain J ( p − / ≡ − (mod p ).It is known that Gaussian hypergeometric series (see, for example, [3–6, 12]) and theW-Z method (see, for example, [2, 8, 11, 15]) commonly apply to the Ramanujan-typesupercongruences. We refer to [8, 12] for more recent developments on van Hamme’s su-percongruences. In this paper, we provide a different approach which is based on somecombinatorial identities involving harmonic numbers. All of these identities are automat-ically discovered and proved by the software package Sigma developed by Schneider [9].The rest of this paper is organized as follows. The preliminary section, Section 2, isdevoted to some properties of the p -adic Gamma function. We prove Theorems 1.3 and1.5 in Sections 3 and 4, respectively. We first recall the definition and some basic properties of the p -adic Gamma function.For more details, we refer to [1, Section 11.6]. Let p be an odd prime and Z p denote theset of all p -adic integers. For x ∈ Z p , the p -adic Gamma function is defined asΓ p ( x ) = lim m → x ( − m Y In order to prove Theorem 1.3, we establish the following two combinatorial identities. Lemma 3.1 For any odd integer n , we have n X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k ( − n ) k ( n + 1) k (1) k ( n + ) k ( − n + ) k = 0 . (3.1) Proof. We begin with the following identity [6, (2.1)]: F " a, a + 1 , b, c, d, e a , a − b + 1 , a − c + 1 , a − d + 1 , a − e + 1; − = Γ( a − d + 1)Γ( a − e + 1)Γ( a + 1)Γ( a − d − e + 1) F " a − b − c + 1 , d, ea − b + 1 , a − c + 1; 1 . Letting a = d = e = , b = − n and c = n + 1 in the above gives n X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k ( − n ) k ( n + 1) k (1) k ( n + ) k ( − n + ) k = F " , , n + , − n + ; 1 Γ (cid:18) (cid:19) Γ (cid:18) (cid:19) . (3.2)Recall the following Whipple’s identity [10, page 54]: F " a, − a, ce, c − e ; 1 = 2 − c π Γ( e )Γ(1 + 2 c − e )Γ (cid:0) e + a (cid:1) Γ (cid:0) + e − a (cid:1) Γ (cid:0) + c − e + a (cid:1) Γ (cid:0) c − e − a (cid:1) . (3.3)Assume that n is odd. Letting a = c = and e = n + in (3.3) and noting that both1 − ( n + 1) and (1 − n ) are non-positive integers, we conclude that F " , , n + , − n + ; 1 = 0 . (3.4)Combining (3.2) and (3.4), we complete the proof of (3.1). (cid:3) Lemma 3.2 For any odd integer n , we have n X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k ( − n ) k ( n + 1) k (1) k ( n + ) k ( − n + ) k k X j =1 (cid:18) j ) − j − (cid:19) = 4 n − (2 n + 1) (cid:30) n (cid:18) n − n − (cid:19) . (3.5)4 roof. Actually, we can automatically discover and prove (3.5) by use of the softwarepackage Sigma developed by Schneider (see [7, Section 5] and [9, Section 3.1] for a similarapproach to finding and proving identities of this type).After loading Sigma into Mathematica, we insert: In[1] := mySum = n +1 X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k ( − n − k (2 n + 2) k (1) k (2 n + ) k ( − n − ) k k X j =1 (cid:18) j ) − j − (cid:19) We compute the recurrence for the above sum: In[2] := rec = GenerateRecurrence[mySum , n ][[1]]We find that the above sum SUM[ n ] satisfies a recurrence of order 2. Now we solve thisrecurrence: In[3] := recSol = SolveRecurrence[rec , SUM[ n ]] Out[3] = , − n )(1 + 2 n ) n Y ι =1 ι ( − ι ) , , n (1 + 2 n ) n Y ι =1 ι ( − ι ) n X ι =1 ι − n X ι =1 ι ) , { , } Finally, we combine the solutions to represent mySum as follows: In[4] := FindLinearCombination[recSol , mySum , Out[4] = 3 + 4 n n ) n Y ι =1 ι ( − ι ) This implies that both sides of (3.5) are equal. (cid:3) Proof of Theorem 1.3. Assume that p ≡ n = p − in (3.1) yields p − X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k ( − p ) k ( p ) k (1) k (1 + p ) k (1 − p ) k = 0 . (3.6)We next determine the following product modulo p :( − p ) k ( p ) k (1 + p ) k (1 − p ) k = k Y j =1 (2 j − − p (2 j ) − p . From the following two Taylor expansions:(2 j − − x (2 j ) − x = (cid:18) j − j (cid:19) − j − j ) x + O ( x ) , and k Y j =1 ( a j + b j x ) = k Y j =1 a j · x k X j =1 b j a j ! + O ( x ) , 5e deduce that for 0 ≤ k ≤ p − ,( − p ) k ( p ) k (1 + p ) k (1 − p ) k ≡ k Y j =1 (cid:18) j − j (cid:19) − (4 j − p (2 j ) ! ≡ (cid:0) (cid:1) k (1) k ! p k X j =1 (cid:18) j ) − j − (cid:19)! (mod p ) . (3.7)Substituting (3.7) into (3.6) gives p − X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k k ! ! ≡ − p p − X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k k ! ! k X j =1 (cid:18) j ) − j − (cid:19) (mod p ) . (3.8)Furthermore, letting n = p − in (3.5) and noting the fact that( a + bp ) k ( a − bp ) k ≡ ( a ) k (mod p ) , (3.9)we obtain p − X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k k ! ! k X j =1 (cid:18) j ) − j − (cid:19) ≡ p p − (cid:30) (cid:18) p − (cid:19) (cid:18) p − p − (cid:19) (mod p ) . (3.10)It follows from (3.8) and (3.10) that p − X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k k ! ! ≡ − p p − (cid:30) (cid:18) p − (cid:19) (cid:18) p − p − (cid:19) (mod p ) . (3.11)By the Fermat’s little theorem, we have2 p − (cid:30) (cid:18) p − (cid:19) ≡ 14 (mod p ) . (3.12)On the other hand, using (2.1), (2.4) and (2.5) we obtain (cid:18) p − p − (cid:19) = 2 p − (cid:0) (cid:1) p − (1) p − = 2 p − Γ p (cid:0) p − (cid:1) Γ p (1) Γ p (cid:0) (cid:1) Γ p (cid:0) p +14 (cid:1) ≡ · Γ p (cid:0) − (cid:1) Γ p (cid:0) (cid:1) Γ p (cid:0) (cid:1) (mod p ) . (3.13)6rom (3.11)–(3.13), we deduce that p − X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k k ! ! ≡ − p Γ p (cid:0) (cid:1) Γ p (cid:0) (cid:1) Γ p (cid:0) − (cid:1) (mod p ) . Since p ≡ p (cid:18) (cid:19) = ( − p +12 = 1 and Γ p (cid:18) − (cid:19) Γ p (cid:18) (cid:19) = 1 . It follows that p − X k =0 ( − k (4 k + 1) (cid:0) (cid:1) k k ! ! ≡ − p Γ p (cid:18) (cid:19) Γ p (cid:18) (cid:19) (mod p ) . (3.14)Finally, by (2.3) we obtain Γ p (cid:18) (cid:19) = − 14 Γ p (cid:18) (cid:19) . Substituting the above into (3.14), we complete the proof of (1.3). (cid:3) Lemma 4.1 For any odd integer n , we have n X k =0 (cid:0) (cid:1) k ( − n ) k ( n + 1) k (1) k = 0 . (4.1) Proof. Assume that n is odd. Letting a = − n, e = 1 and c = in (3.3) and noting that − n is a non-positive integer, we immediately arrive at (4.1). (cid:3) Lemma 4.2 For any odd integer n , we have n X k =0 (cid:0) (cid:1) k ( − n ) k ( n + 1) k (1) k k X j =1 j − = − n − n − X k =0 (cid:0) kk (cid:1) k (cid:30) n (cid:18) n − n − (cid:19) . (4.2) Proof. The above identity possesses, of course, the same automated proof as (3.5), andwe omit the details. (cid:3) Proof of Theorem 1.5. Assume that p ≡ n = p − in (4.1), we obtain p − X k =0 (cid:0) (cid:1) k (cid:0) p (cid:1) k (cid:0) − p (cid:1) k (1) k = 0 . (4.3)7imilarly to the proof of (3.7), we can show that for 0 ≤ k ≤ p − , (cid:0) p (cid:1) k (cid:0) − p (cid:1) k (1) k ≡ (cid:0) (cid:1) k (1) k ! − p k X j =1 j − ! (mod p ) . (4.4)Substituting (4.4) into (4.3) gives p − X k =0 (cid:0) (cid:1) k k ! ! ≡ p p − X k =0 (cid:0) (cid:1) k k ! ! k X j =1 j − (mod p ) . (4.5)Furthermore, letting n = p − in (4.2) and then using (3.9), we arrive at p − X k =0 (cid:0) (cid:1) k k ! ! k X j =1 j − ≡ − p − J p − / (cid:30) (cid:18) p − (cid:19) (cid:18) p − p − (cid:19) (mod p ) . (4.6)By (2.4), we have4 p − (cid:30) (cid:18) p − (cid:19) (cid:18) p − p − (cid:19) = 14 · (1) p − (cid:0) (cid:1) p +14 = 14 · Γ p (cid:0) p +14 (cid:1) Γ p (cid:0) (cid:1) Γ p (1) Γ p (cid:0) p +34 (cid:1) . (4.7)Since p ≡ p (1) = 1 , Γ p (cid:18) (cid:19) = ( − p +12 = 1 , Γ p (cid:18) p + 34 (cid:19) Γ p (cid:18) − p (cid:19) = 1 . Substituting the above into (4.7) yields4 p − (cid:30) (cid:18) p − (cid:19) (cid:18) p − p − (cid:19) = 14 Γ p (cid:18) p (cid:19) Γ p (cid:18) − p (cid:19) . By (2.5), we have Γ p (cid:18) p (cid:19) ≡ Γ p (cid:18) (cid:19) (cid:18) p G (cid:18) (cid:19)(cid:19) (mod p ) , Γ p (cid:18) − p (cid:19) ≡ Γ p (cid:18) (cid:19) (cid:18) − p G (cid:18) (cid:19)(cid:19) (mod p ) . 8t follows that Γ p (cid:18) p (cid:19) Γ p (cid:18) − p (cid:19) ≡ Γ p (cid:18) (cid:19) (mod p ) , and so 4 p − (cid:30) (cid:18) p − (cid:19) (cid:18) p − p − (cid:19) ≡ 14 Γ p (cid:18) (cid:19) (mod p ) . (4.8)Then the proof of (1.5) follows from (4.5), (4.6) and (4.8). 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